13" 5lb TITIANIUM brake rotors??!?!
#1
13" 5lb TITIANIUM brake rotors??!?!
8000 bucks but still..... FIVE pounds.....
http://www.bavauto.com/se1.asp?dept_id=216#top
http://www.bavauto.com/se1.asp?dept_id=216#top
#4
how much lighter are the ceramic ones on the porsches?
hmm... yea stickier tires would be a better investment compared to 8000 USD lol but just the first time I've seen titianum brake rotors myself
hmm... yea stickier tires would be a better investment compared to 8000 USD lol but just the first time I've seen titianum brake rotors myself
#6
Originally posted by savedsol
No matter how sticky your tires are, it doesn't matter much when your brakes stop working from overheating.
No matter how sticky your tires are, it doesn't matter much when your brakes stop working from overheating.
There are very few instance, other than track racing, where I will overheat my brakes.
#7
That's still an uber cool mod. I can hear it now, "Yes, those are titanium rotors on my titanium gray RX-8..." Ok, it's a silly price for a mod, but that is really cool. I doubt I could overheat my factory brake system unless I was actually trying to (and maybe not even then), but I like the look.
#8
The 5 lb part has nothing to do with fade resistance. It's about reducing unsprung and rotating mass, which will help your car accelerate, corner, and stop better. Yes, get the good tires first, but if you really want to wring as much as possible out of those tires, 5 lb rotors would sure help.
They list a 12 lb weight reduction for an M3. That's some serious unsprung mass reduction.
They list a 12 lb weight reduction for an M3. That's some serious unsprung mass reduction.
#9
5-lbfs. reduction in unsprung weight gives me what improvement in the car's performance?
If it's shaving a couple of seconds of off lap time, then I would say spend the money on tires and high performance driving classes and still have money left over. For that kind of $$$ I need to see significant improvement (10's of seconds in improvement)
If it's shaving a couple of seconds of off lap time, then I would say spend the money on tires and high performance driving classes and still have money left over. For that kind of $$$ I need to see significant improvement (10's of seconds in improvement)
#14
Originally posted by Rotary Titus
they say whatever you shave off unsprung weight, it translates to roughly 10xs the weight of your car.
i.e you shave 10 lbs off unsprung = shaving 100lbs off your car.
they say whatever you shave off unsprung weight, it translates to roughly 10xs the weight of your car.
i.e you shave 10 lbs off unsprung = shaving 100lbs off your car.
#16
Reducing weight at the rotors (or wheels) removes two kinds of weight: Rotating and unsprung. They are different concepts, and impact the performance of the car in different ways.
Reducing rotating weight helps the car accelerate and brake faster in a similar fashion that "regular" weight does - the car is simply lighter. The difference is, since the weight is rotating it counts for "more" than it would if it were not rotating. How much more? That depends on how far the weight is from the center of the wheel. If all the weight is located at the periphery of the wheel, such as the tread of the tire, it counts for 2x. If it is at the center of the wheel, it counts for 1x. In between, like a brake rotor, is, well, more than 1x but less than 2x. So, in this case 4 rotors that save 12 lbs each will be like taking, say 75 lbs or so out of the car.
Now, unsprung weight is another matter completely, and it is generally considered more important than rotating weight, pound for pound.
The "unsprung" weight at any corner is the total weight of all the stuff that goes up and down with the wheel. So, it's the wheel, the brake rotor, the brake caliper & pads, the suspension upright, the bottom of the shock, and other bits as well.
The reason unsprung weight is important is that the ability of the car's suspension to keep the tire firmly planted to the ground is directly proportional to the unsprung weight at that corner. It's the shock's job to keep vertical wheel movement to a minimum - the more the wheel moves relative to the road, the more the sidewall flexes veritcally as you drive through a curve, and the less grip you have. The more unsprung weight you have, the harder it is for the shock to control this movement.
Engineering types call this vertical sidewall flex "tire force variation" because it literally results in variations in the tire's reaction force with the road (and remember, friction between the tire and road is proportional to the tire's reaction force) and they aim to minimize it. Top-tier racing teams spend gobs of money to minimize it. Every pound of unsprung weight added results in a reduction in grip over anything less than a perfect road surface.
So, in the case of the $8000 dollar brake upgrade, you save 12 lbs from the rotor and 5 lbs from the caliper. That's a 17 lb unsprung weight reduction, which is huge. Now, is it worth $8000? Not unless you already have the absolute lightest wheels money can buy and tires to match (and money to burn). But I can gaurentee that anybody who likes to drive their RX-8 fast over real-world roads would immediately notice the difference with a 17 lb reduction in unsprung mass. It's huge. I can feel 4 lbs in my Miata easily, and am contemplating spending $$$ to save 6-8 lbs.
So, yes, in the hands of a good driver, these brakes would reduce lap times, even if they didn't help braking performance directly.
Reducing rotating weight helps the car accelerate and brake faster in a similar fashion that "regular" weight does - the car is simply lighter. The difference is, since the weight is rotating it counts for "more" than it would if it were not rotating. How much more? That depends on how far the weight is from the center of the wheel. If all the weight is located at the periphery of the wheel, such as the tread of the tire, it counts for 2x. If it is at the center of the wheel, it counts for 1x. In between, like a brake rotor, is, well, more than 1x but less than 2x. So, in this case 4 rotors that save 12 lbs each will be like taking, say 75 lbs or so out of the car.
Now, unsprung weight is another matter completely, and it is generally considered more important than rotating weight, pound for pound.
The "unsprung" weight at any corner is the total weight of all the stuff that goes up and down with the wheel. So, it's the wheel, the brake rotor, the brake caliper & pads, the suspension upright, the bottom of the shock, and other bits as well.
The reason unsprung weight is important is that the ability of the car's suspension to keep the tire firmly planted to the ground is directly proportional to the unsprung weight at that corner. It's the shock's job to keep vertical wheel movement to a minimum - the more the wheel moves relative to the road, the more the sidewall flexes veritcally as you drive through a curve, and the less grip you have. The more unsprung weight you have, the harder it is for the shock to control this movement.
Engineering types call this vertical sidewall flex "tire force variation" because it literally results in variations in the tire's reaction force with the road (and remember, friction between the tire and road is proportional to the tire's reaction force) and they aim to minimize it. Top-tier racing teams spend gobs of money to minimize it. Every pound of unsprung weight added results in a reduction in grip over anything less than a perfect road surface.
So, in the case of the $8000 dollar brake upgrade, you save 12 lbs from the rotor and 5 lbs from the caliper. That's a 17 lb unsprung weight reduction, which is huge. Now, is it worth $8000? Not unless you already have the absolute lightest wheels money can buy and tires to match (and money to burn). But I can gaurentee that anybody who likes to drive their RX-8 fast over real-world roads would immediately notice the difference with a 17 lb reduction in unsprung mass. It's huge. I can feel 4 lbs in my Miata easily, and am contemplating spending $$$ to save 6-8 lbs.
So, yes, in the hands of a good driver, these brakes would reduce lap times, even if they didn't help braking performance directly.
#17
Well, after a little research, I did not find any support of the 2for1 reduction I mentioned earlier. I did find one mention of 10for1, but is was in a Honda forum not unlike this one, so I can't speak as to it's validity. Also, I did find on a drag racer forum that a 100lb weight reduction is generally equivalent to a 0.1 second reduction in 1/4 mile e.t. All this being said, assuming we could save 17lbs from each corner and apply the 10x factor, we shave nearly seven tenths of of the 1/4. While there are a host of less expensive options to reduce 1/4 mile times, if you are a money-is-no-object kind of guy which not many of us are (wouldn't you just pay someone to read the forum for you) it would definetely be worth considering.
#19
Originally posted by Rotary Titus
they say whatever you shave off unsprung weight, it translates to roughly 10xs the weight of your car.
i.e you shave 10 lbs off unsprung = shaving 100lbs off your car.
they say whatever you shave off unsprung weight, it translates to roughly 10xs the weight of your car.
i.e you shave 10 lbs off unsprung = shaving 100lbs off your car.
Handling would be hard to calculate, but I have done several times for acceleration and it is more like 1.5x not 10x.
From an earlier post of mine:
The kinetic energy of a non-rotational mass experiencing translational movement is:
Kt = 1/2 *mv^2,
where m is mass in Kg and v is velocity in m/s. For a rotating mass experiencing translational movement, the kinetic energy is:
Krt = 1/2 * mv^2 + 1/2 * Iw^2
where I is the moment of inertia and w is the angular velocity in radians per second. For a 18 inch (.4572 meter) wheel that has all of its mass on the outer rim (worstcase), it has an I of:
I = mr^2 = .0523m
and has a angualar velocity of:
w = 2pi * v / (2pi * r) = 3.03v
where r here is the radius of the 26" wide tire. Substituting back in to the earlier equation we have:
Krt = 1/2 * mv^2 + 1/2 *.0523m * (3.03v)^2 = .7398 mv^2
so Krt = .7398/.5 Kt = 1.480 Kt
Or, stated differently the energy needed to accelerate every 1 lb of wheel weight of an 18 inch wheel is equal to 1.48 lbs of payload weight in the car. I don't know where the "4 times" number comes from.
Kt = 1/2 *mv^2,
where m is mass in Kg and v is velocity in m/s. For a rotating mass experiencing translational movement, the kinetic energy is:
Krt = 1/2 * mv^2 + 1/2 * Iw^2
where I is the moment of inertia and w is the angular velocity in radians per second. For a 18 inch (.4572 meter) wheel that has all of its mass on the outer rim (worstcase), it has an I of:
I = mr^2 = .0523m
and has a angualar velocity of:
w = 2pi * v / (2pi * r) = 3.03v
where r here is the radius of the 26" wide tire. Substituting back in to the earlier equation we have:
Krt = 1/2 * mv^2 + 1/2 *.0523m * (3.03v)^2 = .7398 mv^2
so Krt = .7398/.5 Kt = 1.480 Kt
Or, stated differently the energy needed to accelerate every 1 lb of wheel weight of an 18 inch wheel is equal to 1.48 lbs of payload weight in the car. I don't know where the "4 times" number comes from.
On a really light car like the Miata (and the RX-8 to a lesser extent) I can see why every once matters. But if weight is so important why do the S3 tires from Bridgestone do so well in comparison tests at Tire Rack when they are sometimes as much as 4-5 pounds heavier than the competition? I'm talking track times here on a 3-series BMW (i.e. about the same car as the RX-8), not to mention this is weight at the very outer edge of the tire (now that is getting closer to the 2x range).
No, you don't want 60lb Kriminals on your car, but shaving so much weight to compromise structural integrity of your tire/rim combo might also be diminishing returns and that is why the S3 tires perform so well. Light tires that have overly plyable sidewalls and light potentially flemsy rims that contort under heavy corning might give up their weight benefits for other disadvantages.
-Mr. Wigggles
Last edited by MrWigggles; 04-16-2004 at 02:59 AM.
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